Inquiry involving calculating Galois group of splitting field of $(x^2-3)(x^3-3).$ over $\mathbb{Q}.$

Question

I factor the polynomial as $(x-\sqrt{3})(x+\sqrt{3})(x-\sqrt[3]{3})(x-\rho\sqrt[3]{3})(x-\rho^2\sqrt[3]{3})$ where $\rho$ is the primitive 3rd root of unity. The splitting field will be $\mathbb{Q}(\sqrt{3}, \sqrt[3]{3}, \rho\sqrt[3]{3}).$ When it comes to writing down the Galois group I run into a problem. I have two automorphisms which fix $\mathbb{Q}$, $$\sigma:\bigg(\sqrt[3]{3} \mapsto \rho\sqrt[3]{3}, \ \rho \to \rho \bigg)$$ $$\tau:\bigg(\sqrt[3]{3} \mapsto \sqrt[3]{3}, \ \rho \to \rho^2 \bigg).$$

My question is whether these maps should also fix $\sqrt{3}?$ Because $\rho = (-1+\sqrt{-3})/2$ and $\rho^2 = (-1-\sqrt{-3})/2$, $\tau(\sqrt{-3}) = -\sqrt{-3}.$ Does this necessarily imply that $\tau(\sqrt{3}) = -\sqrt{3}$ and thus permutes the first two roots above? Or does $\tau(i) = -i$ and hence fixes $\sqrt{3}$?

Answer 1

$\tau(i\sqrt 3)=-\tau(i\sqrt 3)$ doesn't really imply anything about $\tau(\sqrt 3)$ because there's no way to algebraically express $\sqrt 3$ in terms of adding/multiplying/squaring $i\sqrt 3$. This can be shown more formally by proving that $x^2-3$ has no roots in $\Bbb{Q}(i\sqrt 3)$: Any $x \in \Bbb{Q}(i\sqrt 3)$ is in the form of $x=a+bi\sqrt 3$ because $i\sqrt 3$ has minimal degree $2$. However, when you square this, you get $x^2=a^2-3b^2+2ab\sqrt 3i$. If we want to solve $x^2=3$, we need a zero imaginary part, which happens only if $a=0$ or $b=0$. If $b=0$, then $x$ is rational, so $x^2 \neq 3$. If $a=0$, then $x$ is imaginary, so $x^2<0\rightarrow x^2\neq 3$. Thus, $x^2\neq 3$ for all $x\in \Bbb{Q}(i\sqrt 3)$.

Therefore, $\tau(\sqrt 3)$ is not determined by $\tau(i\sqrt 3)$, so it could or could not permute the roots, depending on what you choose.

Answer 2

The splitting field over $\mathbf Q$ of $(X^2-3)(X^3-3)$ is obviously $L=\mathbf Q (\sqrt 3, \sqrt [3]3, \omega)$, where $\omega$ is a primitive cubic root of unity. Then the description of $\mathcal G=Gal(L/\mathbf Q)$ is just a question of organizing the relevant subgroups/quotients of $\mathcal G$. The most convenient way seems to introduce $K=\mathbf Q (\sqrt [3]3, \omega)$ and write $L=K(\sqrt 3)$. The group $Gal(K/\mathbf Q)\cong S_3$ admits a unique quotient of order $2$, hence a unique quadratic subfield which is $\mathbf Q (\sqrt {-3})$, and so $K$ and $k=\mathbf Q(\sqrt 3)$ are linearly disjoint over $\mathbf Q$. It follows that $\mathcal G$ is a direct product, more pecisely the direct product of the subgroups $Gal(L/K) \cong Gal(k/\mathbf Q) \cong C_2$ and $Gal (L/k)\cong Gal(K/\mathbf Q)\cong S_3$. It remains only to make the generators explicit. The generator $\sigma$ of $Gal(L/K)$ is determined by $\sigma (\sqrt 3)=-\sqrt 3$ (the other field generators $\sqrt [3]3, \omega$ being automatically invariant), whereas those of $Gal (L/k)$ are the transposition $\tau:\omega \to \omega^2$ and the $3$-cycle $\gamma: \sqrt [3]3 \to \omega \sqrt [3]3$ (the other non-mentioned field generators being automatically invariant).